ECE 280/Examples/Convolution

The following is an example of convolving two signals; the convolution is done several different ways:

Math... So much math.
Using Convolution Shortcuts
Geometrically, flipping and shifting $$ h(t) $$
Geometrically, flipping and shifting $$ x(t) $$

Setup

The goal for this problem is to determine the output $$ y(t) $$ created by an input $$ x(t) $$ for a linear time invariant system if the system's impulse response $$ h(t) $$ is known. In this particular example:

\begin{align} x(t)&=2u(t)-u(t-1)-u(t-3)\\ h(t)&=e^{-t}u(t)\\ y(t)&=x(t)*h(t)=\int_{-\infty}^{\infty}x(\tau)h(t-\tau)d\tau=\int_{-\infty}^{\infty}x(t-\tau)h(\tau)d\tau \end{align}

where the particular choice of which integral to use is up to the user.

Math

Given the relative complexity of $$ x(t) $$ , it may make more sense to use the first form, where $$ h(t) $$ is flipped and shifted. That yields:

\begin{align} y(t)&=\int_{-\infty}^{\infty}x(\tau)h(t-\tau)d\tau\\ y(t)&=\int_{-\infty}^{\infty}\left(2u(\tau)-u(\tau-1)-u(\tau-3)\right)\left(e^{-(t-\tau)}u(t-\tau)\right)d\tau \end{align}

Distributing terms gives:

\begin{align} y(t)&=\int_{-\infty}^{\infty}2e^{-(t-\tau)}u(\tau)u(t-\tau)d\tau\dots\\ ~&~-\int_{-\infty}^{\infty}e^{-(t-\tau)}u(\tau-1)u(t-\tau)d\tau\dots\\ ~&~-\int_{-\infty}^{\infty}e^{-(t-\tau)}u(\tau-3)u(t-\tau)d\tau\\ \end{align}

Taking the first part alone:

\int_{-\infty}^{\infty}{\color{Brown}2e^{-(t-\tau)}}{\color{Red}u(\tau)}{\color{Blue}u(t-\tau)}d\tau

note that the integrand is only non-zero when two conditions are met simultaneously:

\begin{align} {\color{Red}0}&{\color{Red}<\tau} & {\color{Blue}\tau}&{\color{Blue}<t} \end{align}

Furthermore, those conditions can only be met simultaneously if the outer conditions are met; that is, if:

\begin{align} {\color{Purple}0}&{\color{Purple}<t} \end{align}

These three conditions lead to three changes. The limits on $\tau$ change the limits of the integral with respect to $\tau$ while the condition on $$ t $$ can be represented by a unit step of $$ t $$ . In other words:

\begin{align} \int_{-\infty}^{\infty}{\color{Brown}2e^{-(t-\tau)}}{\color{Red}u(\tau)}{\color{Blue}u(t-\tau)}d\tau &= {\color{Purple}u(t)}\int_{{\color{Red}0}}^{{\color{Blue}t}}{\color{Brown}2e^{-(t-\tau)}}d\tau\\ {\color{Purple}u(t)}\int_{{\color{Red}0}}^{{\color{Blue}t}}{\color{Brown}2e^{-(t-\tau)}}d\tau&= u(t)\left[2e^{-(t-\tau)} \right]_0^{t}=\left(2-2e^{-t}\right)u(t) \end{align}

Note with that last line that the integral is with respect to $\tau$ and the exponent of $\tau$ is +1. The other two components of the convolution integral can be processed in a similar way:

\begin{align} \int_{-\infty}^{\infty}{\color{Brown}-e^{-(t-\tau)}}{\color{Red}u(\tau-1)}{\color{Blue}u(t-\tau)}d\tau &= {\color{Purple}u(t-1)}\int_{{\color{Red}1}}^{{\color{Blue}t}}{\color{Brown}-e^{-(t-\tau)}}d\tau\\ {\color{Purple}u(t-1)}\int_{{\color{Red}1}}^{{\color{Blue}t}}{\color{Brown}-e^{-(t-\tau)}}d\tau&= -u(t-1)\left[e^{-(t-\tau)} \right]_1^{t}=-\left(1-1e^{-(t-1)}\right)u(t-1) \end{align}

and

\begin{align} \int_{-\infty}^{\infty}{\color{Brown}-e^{-(t-\tau)}}{\color{Red}u(\tau-3)}{\color{Blue}u(t-\tau)}d\tau &= {\color{Purple}u(t-3)}\int_{{\color{Red}3}}^{{\color{Blue}t}}{\color{Brown}-e^{-(t-\tau)}}d\tau\\ {\color{Purple}u(t-3)}\int_{{\color{Red}3}}^{{\color{Blue}t}}{\color{Brown}-e^{-(t-\tau)}}d\tau&= -u(t-3)\left[e^{-(t-\tau)} \right]_3^{t}=-\left(1-1e^{-(t-3)}\right)u(t-3) \end{align}

meaning (finally) that:

\begin{align} y(t)=x(t)*h(t)=\left(2-2e^{-t}\right)u(t)-\left(1-1e^{-(t-1)}\right)u(t-1)-\left(1-1e^{-(t-3)}\right)u(t-3) \end{align}

Written piecewise, this equation would be:

\begin{align} y(t)&=\begin{cases} {\color{Black}0} & {\color{Black}t<0} \\ 2-2e^{-t} & {\color{Black}0<t<1}\\ 1 - 2e^{-t}+ e^{-(t-1)} & {\color{Black}1<t<3}\\ -2e^{-t} + e^{-(t-1)} + e^{-(t-3)} & {\color{Black}t>3} \end{cases} \end{align}

Shortcuts

Full color version:

To use the shortcuts, first distribute the various components of the convolution:

\begin{align} x(t)&=\color{Brown}{2u(t)-u(t-1)-u(t-3)}\\ h(t)&=\color{Blue}{e^{-t}u(t)}\\ x(t)*h(t)&=\color{Brown}{2u(t)}*\color{Blue}{e^{-t}u(t)}\color{Brown}{-u(t-1)}*\color{Blue}{e^{-t}u(t)}\color{Brown}{-u(t-3)}*\color{Blue}{e^{-t}u(t)} \end{align}

These can be rearranged, pulling out any shifts, as:

\begin{align} x(t)*h(t)&=\color{Orange}{2}\color{Blue}{e^{-t}u(t)}*\color{Red}{u(t)}\color{Orange}{-}\color{Blue}{e^{-t}u(t)}*\color{Red}{u(t)}\color{Orange}{*\delta(t-1)}\color{Orange}{-}\color{Blue}{e^{-t}u(t)}*\color{Red}{u(t)}\color{Orange}{*\delta(t-3)} \end{align}

Note that each component now has a convolution of

\begin{align} \color{Blue}{e^{-t}u(t)}*\color{Red}{u(t)} \end{align}

in it. Convolution with a step generates an integral, and integrating the right-sided exponential yields:

\begin{align} \color{Red}{\int_{-\infty}^{t}}\color{Blue}{e^{-\tau}u(\tau)}d\tau=u(t)\left[-e^{-\tau}\right]_0^t=\color{Fuchsia}{(1-e^{-t})u(t)} \end{align}

which means the rearranged functions can be re-written as:

\begin{align} x(t)*h(t)&=\color{Orange}{2}\color{Fuchsia}{(1-e^{-t})u(t)}\color{Orange}{-}\color{Fuchsia}{(1-e^{-t})u(t)}\color{Orange}{*\delta(t-1)}\color{Orange}{-}\color{Fuchsia}{(1-e^{-t})u(t)}\color{Orange}{*\delta(t-3)} \end{align}

Now the shifts can be applied to give:

\begin{align} x(t)*h(t)&=2(1-e^{-t})u(t)-(1-e^{-(t-1)})u(t-1)-(1-e^{-(t-3)})u(t-3) \end{align}

Black and white only:

To use the shortcuts, first distribute the various components of the convolution:

\begin{align} x(t)&={\color{Black}2u(t)-u(t-1)-u(t-3)}\\ h(t)&={\color{Black}e^{-t}u(t)}\\ x(t)*h(t)&={\color{Black}2u(t)}*{\color{Black}e^{-t}u(t)}{\color{Black}-u(t-1)}*{\color{Black}e^{-t}u(t)}{\color{Black}-u(t-3)}*{\color{Black}e^{-t}u(t)} \end{align}

These can be rearranged, pulling out any shifts, as:

\begin{align} x(t)*h(t)&={\color{Black}2}{\color{Black}e^{-t}u(t)}*{\color{Black}u(t)}{\color{Black}-}{\color{Black}e^{-t}u(t)}*{\color{Black}u(t)}{\color{Black}*\delta(t-1)}{\color{Black}-}{\color{Black}e^{-t}u(t)}*{\color{Black}u(t)}{\color{Black}*\delta(t-3)} \end{align}

Note that each component now has a convolution of

\begin{align} {\color{Black}e^{-t}u(t)}*{\color{Black}u(t)} \end{align}

in it. Convolution with a step generates an integral, and integrating the right-sided exponential yields:

\begin{align} {\color{Black}\int_{-\infty}^{t}}{\color{Black}e^{-\tau}u(\tau)}d\tau=u(t)\left[-e^{-\tau}\right]_0^t={\color{Black}(1-e^{-t})u(t)} \end{align}

which means the rearranged functions can be re-written as:

\begin{align} x(t)*h(t)&={\color{Black}2}{\color{Black}(1-e^{-t})u(t)}{\color{Black}-}{\color{Black}(1-e^{-t})u(t)}{\color{Black}*\delta(t-1)}{\color{Black}-}{\color{Black}(1-e^{-t})u(t)}{\color{Black}*\delta(t-3)} \end{align}

Now the shifts can be applied to give:

\begin{align} x(t)*h(t)&=2(1-e^{-t})u(t)-(1-e^{-(t-1)})u(t-1)-(1-e^{-(t-3)})u(t-3) \end{align}

Geometrically Flipping/Shifting h

The geometric method involves determining the piecewise regions in $\tau$ where the different functions overlap as a function of $$ t $$ . With the current example, the only change $$ h(t) $$ makes is at $$ t=0 $$ , while $$ x(t) $$ has four different parts. The piecewise definitions of the two functions can be written as:

\begin{align} x(t)&= \begin{cases} 0 & t<0 \\ 2 & 0<t<1 \\ 1 & 1<t<3 \\ 0 & t>3\end{cases} & h(t)&= \begin{cases} 0 & t<0 \\ e^{-t} & t>0\end{cases} & \end{align}

In the convolution integral

\begin{align} y(t)&=x(t)*h(t)=\int_{-\infty}^{\infty}x(\tau)h(t-\tau)d\tau \end{align}

the piecewise definitions of the signals change to functions of $\tau$ :

\begin{align} x(\tau)&= \begin{cases} 0 & \tau<0 \\ 2 & 0<\tau<1 \\ 1 & 1<\tau<3 \\ 0 & t>3\end{cases} & h(t-\tau)&= \begin{cases} e^{-(t-\tau)} & \tau<t \\ 0 & \tau>t\end{cases} \end{align}

As $$ t $$ changes, $x(\tau)$ does not move but $h(t-\tau)$ does, resulting in different components of the two functions overlapping at different times. The following four subsections will show what happens as $$ t $$ increases.

t<0

The "leading edge" of $h(t-\tau)$ is at $\tau=t$ ; until $$ t>0 $$ , there will be no overlap with $x(\tau)$ and thus the convolution integral will be 0:

\begin{align} y(t)&=\begin{cases} {\color{Black}0} & {\color{Black}t<0} \end{cases} \end{align}

0<t<1

Once $$ t>0 $$ , the segment of $$ h $$ that is nonzero passes into the region of $$ x $$ that is equal to the constant 2. This will remain the situation until $$ h $$ 's transition into the region where $$ x $$ is 1. During this phase, the integrand will be the product of the exponential from $$ h $$ and the 2 from $$ x $$ :

\begin{align} y(t)&=\begin{cases} {\color{Black}0} & {\color{Black}t<0} \\ {\color{Black}\int_{{\color{Red}0}}^{{\color{Blue}t}}{\color{Red}2}{\color{Blue}e^{-(t-\tau)}}~d\tau} & {\color{Black}0<t<1} \end{cases} \end{align}

1<t<3

After $$ t>1 $$ but before $$ t>3 $$ , there are two different regions of overlap: the region where $$ x $$ is 2 now overlaps for $0<\tau<1$ while the region where $$ x $$ is 1 overlaps where $1<\tau<t$ . That yields:

\begin{align} y(t)&=\begin{cases} {\color{Black}0} & {\color{Black}t<0} \\ {\color{Black}\int_{{\color{Red}0}}^{{\color{Blue}t}}{\color{Red}2}{\color{Blue}e^{-(t-\tau)}}~d\tau} & {\color{Black}0<t<1}\\ {\color{Black}\int_{{\color{Red}0}}^{{\color{Red}1}}{\color{Red}2}{\color{Blue}e^{-(t-\tau)}}~d\tau} + {\color{Black}\int_{{\color{Red}1}}^{{\color{Blue}t}}{\color{Red}1}{\color{Blue}e^{-(t-\tau)}}~d\tau} & {\color{Black}1<t<3} \end{cases} \end{align}

t>3

Finally, once $$ t>3 $$ , the leading edge of the flipped $$ h $$ has passed the edge of $$ x $$ and the limits of integration are thus purely dependent on the piecewise boundaries of $$ x $$ alone:

\begin{align} y(t)&=\begin{cases} {\color{Black}0} & {\color{Black}t<0} \\ {\color{Black}\int_{{\color{Red}0}}^{{\color{Blue}t}}{\color{Red}2}{\color{Blue}e^{-(t-\tau)}}~d\tau} & {\color{Black}0<t<1}\\ {\color{Black}\int_{{\color{Red}0}}^{{\color{Red}1}}{\color{Red}2}{\color{Blue}e^{-(t-\tau)}}~d\tau} + {\color{Black}\int_{{\color{Red}1}}^{{\color{Blue}t}}{\color{Red}1}{\color{Blue}e^{-(t-\tau)}}~d\tau} & {\color{Black}1<t<3}\\ {\color{Black}\int_{{\color{Red}0}}^{{\color{Red}1}}{\color{Red}2}{\color{Blue}e^{-(t-\tau)}}~d\tau} + {\color{Black}\int_{{\color{Red}1}}^{{\color{Red}3}}{\color{Red}1}{\color{Blue}e^{-(t-\tau)}}~d\tau} & {\color{Black}t>3} \end{cases} \end{align}

After Integration

After calculating all those integrals, the answer would be:

\begin{align} y(t)&=\begin{cases} {\color{Black}0} & {\color{Black}t<0} \\ 2-2e^{-t} & {\color{Black}0<t<1}\\ (2e^{-(t-1)}-2e^{-t})+(1-e^{-(t-1)}) & {\color{Black}1<t<3}\\ (2e^{-(t-1)}-2e^{-t}) + (e^{-(t-3)}-e^{-(t-1)})& {\color{Black}t>3} \end{cases} \end{align}

or, combined,

\begin{align} y(t)&=\begin{cases} {\color{Black}0} & {\color{Black}t<0} \\ 2-2e^{-t} & {\color{Black}0<t<1}\\ 1 - 2e^{-t}+ e^{-(t-1)} & {\color{Black}1<t<3}\\ -2e^{-t} + e^{-(t-1)} + e^{-(t-3)} & {\color{Black}t>3} \end{cases} \end{align}

Geometrically Flipping/Shifting x

The geometric method involves determining the piecewise regions in $\tau$ where the different functions overlap as a function of $$ t $$ . With the current example, the only change $h(t)$ makes is at $$ t=0 $$ , while $$ x(t) $$ has four different parts. The piecewise definitions of the two functions can be written as:

\begin{align} x(t)&= \begin{cases} 0 & t<0 \\ 2 & 0<t<1 \\ 1 & 1<t<3 \\ 0 & t>3\end{cases} & h(t)&= \begin{cases} 0 & t<0 \\ e^{-t} & t>0\end{cases} & \end{align}

In the convolution integral

\begin{align} y(t)&=x(t)*h(t)=\int_{-\infty}^{\infty}x(t-\tau)h(t)d\tau \end{align}

the piecewise definitions of the signals change to functions of $\tau$ :

\begin{align} x(t-\tau)&= \begin{cases} 0 & \tau<t-3 \\ 1 & t-3<\tau<t-1 \\ 2 & t-1<\tau<0 \\ 0 & \tau>3\end{cases} & h(\tau)&= \begin{cases} 0 & \tau<0 \\ e^{-\tau} & \tau>0\end{cases} \end{align}

As $$ t $$ changes, $h(\tau)$ does not move but $x(t-\tau)$ does, resulting in different components of the two functions overlapping at different times. The following four subsections will show what happens as $$ t $$ increases.

t<0

The "leading edge" of $x(t-\tau)$ is at $\tau=t$ ; until $$ t>0 $$ , there will be no overlap with $h(\tau)$ and thus the convolution integral will be 0:

\begin{align} y(t)&=\begin{cases} {\color{Black}0} & {\color{Black}t<0} \end{cases} \end{align}